You have found the following ages (in years) of all 6 porcupines at your local zoo: $ 4,\enspace 9,\enspace 12,\enspace 2,\enspace 22,\enspace 5$ What is the average age of the porcupines at your zoo? What is the variance? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 porcupines at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{4 + 9 + 12 + 2 + 22 + 5}{{6}} = {9\text{ years old}} $ Find the squared deviations from the mean for each porcupine. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $4$ years $-5$ years $25$ years $^2$ $9$ years $0$ years $0$ years $^2$ $12$ years $3$ years $9$ years $^2$ $2$ years $-7$ years $49$ years $^2$ $22$ years $13$ years $169$ years $^2$ $5$ years $-4$ years $16$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{25} + {0} + {9} + {49} + {169} + {16}} {{6}} $ $ {\sigma^2} = \dfrac{{268}}{{6}} = {44.67\text{ years}^2} $ The average porcupine at the zoo is 9 years old. The population variance is 44.67 years $^2$.